MHB Maximum Likelihood Estimators for Uniform Distribution

[Julio1]

Find maximum likelihood estimators of an sample of size $n$ if $X\sim U(0,\theta].$

Hello MHB :)! Can any user help me please :)! I don't how follow...

 

[I like Serena]

Find maximum likelihood estimators of an sample of size $n$ if $X\sim U(0,\theta].$

Hello MHB :)! Can any user help me please :)! I don't how follow...

Hi Julio!

We want to maximize the likelihood that some $\theta$ is the right one.

The likelihood that a certain $\theta$ is the right one given a random sample is:
$$\mathcal L(\theta; x_1, ..., x_n) = f(x_1|\theta) \times f(x_2|\theta) \times ... \times f(x_n|\theta)$$
where $f$ is the probability density function.

Since $X\sim U(0,\theta]$, $f$ is given by:
$$f(x|\theta)=\begin{cases}\frac 1 \theta&\text{if }0 < x \le \theta \\ 0 &\text{otherwise}\end{cases}$$

Can you tell for which $\theta$ the likelihood will be at its maximum?

 

[Julio1]

Thanks I like Serena :).

Good have that the likelihood function is $L(\theta)=\dfrac{1}{\theta^n}.$ Then applying logarithm we have that

$\ln (L(\theta))=\ln(\dfrac{1}{\theta^n})=-n\ln(\theta).$ Now for derivation with respect $\theta$ we have that $\dfrac{\partial}{\partial \theta}(\ln L(\theta))=\dfrac{\partial}{\partial \theta}(-n\ln(\theta))=-\dfrac{n}{\theta}.$ Thus, match with zero have $-\dfrac{n}{\theta}=0$, i.e., $n=0$.

But why? :(, remove the parameter $\theta$?... Then I don't find an estimator for $\theta$?

 

[I like Serena]

You're welcome Julio!

What's missing in your approach is that it's not taken into account that the function is piecewise.
So we need to inspect what happens at the boundaires.

Note that any $x_i$ that is in the sample has to be $\le \theta$, because otherwise its probability is $0$.
So $\theta$ has to be at least the maximum value that is in the sample.
What happens to the likelihood if $\theta$ is bigger than that maximum value?